If x=2 θ- 2 θ and y=2 θ- 2 θ, then {d² y}{d x²} is equal to

Correct Answer is : 32tan⁡3θ23/2 3 θ/223tan23θ

Correct Answer is : 32tan⁡3θ23/2 3 θ/223tan23θ

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MHTCET 15 May 2024 Shift 1 Paper

Question : 133 of 150

Marks: +1, -0

x=2 \cos \theta-\cos 2 \theta

y=2 \sin \theta-\sin 2 \theta

\frac{d^{2} y}{d x^{2}}

is equal to

Solution:

Concept:

For parametric equations, use

\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}

and then differentiate again with respect to

x

Explanation:

Given

x=2\cos\theta-\cos2\theta

and

y=2\sin\theta-\sin2\theta

Differentiate both with respect to

\theta

\frac{dx}{d\theta}=-2\sin\theta+2\sin2\theta

\frac{dy}{d\theta}=2\cos\theta-2\cos2\theta

Therefore,

\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{\cos\theta-\cos2\theta}{\sin2\theta-\sin\theta}

Using sum-to-product identities:

\cos\theta-\cos2\theta=2\sin\left(\frac{3\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)

\sin2\theta-\sin\theta=2\cos\left(\frac{3\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)

Hence,

\frac{dy}{dx}=\tan\left(\frac{3\theta}{2}\right)

Now,

\frac{d^2y}{dx^2}=\frac{d}{d\theta}\left(\tan\frac{3\theta}{2}\right)\cdot\frac{d\theta}{dx}

=\sec^2\left(\frac{3\theta}{2}\right)\cdot\frac{3}{2}\cdot\frac{1}{2(\sin2\theta-\sin\theta)}

=\frac{3}{4}\left[\frac{\sec^2\left(\frac{3\theta}{2}\right)}{\sin2\theta-\sin\theta}\right]

Answer:

\frac{d^2y}{dx^2}=\frac{3}{4}\left[\frac{\sec^2\left(\frac{3\theta}{2}\right)}{\sin2\theta-\sin\theta}\right]

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